Problem: $\dfrac{d}{dx}[\sin^7(x)]=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $7x^6\cos(x^7)$ (Choice B) B $7\cos^6(x)$ (Choice C) C $\cos(7x^6)$ (Choice D) D $7\sin^6(x)\cos(x)$
Explanation: Since $\sin^7(x)$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. First, we realize that $\sin^7(x)$ is just shorthand for $(\sin(x))^7$, which is clearly composite: $\underbrace{(~\overbrace{\sin(x)}^{\text{inner}}~)^7}_{\text{outer}}$ So if $\sin^7(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={\sin(x)} &&\text{inner function} \\\\ w(x)&=x^7&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={\cos(x)} \\\\ {w'(x)}&={7x^6} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{d}{dx}[\sin^7(x)]&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={7({\sin(x)})^6} \cdot {\cos(x)} \\\\ &=7\sin^6(x)\cos(x) \end{aligned}$